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course PHY 201
10/30 8pm Question 5 and the last little part of question 4 got me a little confused. Can you review and let me know your thoughts? Thanks
`q001. If the acceleration of an Atwood system with total mass 80 grams is 50 cm/s^2, then:* How much mass is on each side? Note that this can be reasoned out easily without an complicated analysis, using the same type of reasoning that led us to conclude that a system with 31 g on one side and 30 g on the other accelerates at 1/61 the acceleration of gravity.
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In this problem we can find the net force of the system and then apply it to the problem.
Using a=Fnet/(m1+M2) and then get a*(m1+m2)=Fnet
Now we have Fnet=50cm/s^2*80g
Fnet=4000g/cm/s^2 or 4N
Applying this net force to the problem, then for both sides to add up to 80g and still have a net force of 4N, then 1 side will be 42g and the other will be 38g. By checking this, we can use the following.
A=Fnet/(M1+M2)
A=(4g*980cm/s^2)/80g=3920g/cm/s^2/80g=49cm/s^2.
The original acceleration was 50cm/s^2 which is what I would have got if I used the acceleration of gravity at 1000cm/s^2
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* By analyzing the forces on the mass on the 'lighter' side, what is the tension in the string?
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The lighter side of the string is 38g
By using Fnet=T-(m g), we get
4N=T-[38g*(980cm/s^2)]
4N=T-(37240g/cm/s^2) or 37N
T=4N+37N
T=41N
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* By analyzing the forces on the mass on the 'heavier' side, what is the tension in the string?
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Before starting this problem, I think that the solution should obviously be more than 41N because the tension of the “lighter side” of the string was 41N. So, let’s see.
The heavier side of the string is 42g
By using Fnet=T-(m g), we get
4N=T-[42g*(980cm/s^2)]
4N=T-(41160g/cm/s^2) or 41N
T=4N+41N
T=45N
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All your reasoning is really good, but your units are causing problems. A kg m/s^2 is 1000 g m/s^2 which is 1000 g * 100 cm / s^2 = 100 000 g cm/s^2. You seem to be associating a Newton with 1000 g cm/s^2 (probably forgetting about the m vs. cm conversion).
You're better off to just keep the units in g, cm and s, and forget about Newtons; force is in g cm/s^2 (that's also called 'dynes' but you don't even need the word). If you do this everything will work out correctly.
`q002. A large sweet potato has mass 1188.6 grams. Sweet potatoes sink in water. Suppose that when this sweet potato is suspended as the mass one one side of an Atwood machine, but immersed in water, a mass of 100 grams on the other side is required to balance it.
* Sketch the forces on that system and describe your sketch.
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On the left side of the sketch is the 100g required to balance the sweet potato. The force of gravity is acting down on the 100g while the tension force is acting up on the 100g.
On the right side of the sketch is the sweet potato submersed in water. There is a buoyant force on the sweet potato that is equal to the amount of water displaced by the sweet potato that is acting down. Also, there is the tension force of the string on that side that is acting up.
Don't forget the gravitational force on the sweet potato.
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* Would the sweet potato, if reshaped without changing its volume, fit into that 1-liter container?
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No, the sweet potato would not fit in the 1-liter container because it’s mass is greater than the 1kg that would fit into the container if that is what the sweet potato weighed.
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* What if the required balancing mass was 200 grams?
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No, I don’t think it would fit. If the balancing mass was 200g, then that would mean the sweet potato would be heavier.
the tension would be 200 g * 980 cm/s^2
the gravitational force would be 1188.6 g * 980 cm/s^2
what therefore must be the buoyant force, and how would it compare with the weight of 1 kg of water?
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`q003. The balloon rose to the ceiling in about 1.5 seconds, 2 seconds, 2.5 seconds and 5 seconds when 1, 2, 3 and 4 paperclips, respectively, were attached. It fell to the floor in about 6 seconds when 5 paperclips were attached. Assume the displacement to have been the same in each case. The buoyant force results from the fact that air pressure decreases as altitude changes, which results in more force from the air pressure on the bottom of the balloon than on the top. The pressure in the room changes at a very nearly constant rate with respect to altitude, so the buoyant force can be assumed to remain constant throughout the room.
* Assuming uniform acceleration in each case, is a graph of acceleration vs. number of clips linear?
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No. The graph was linear until the 4th paper clip was added and then the acceleration spiked up.
Your description appears to be of time interval vs. number of clips, not acceleration vs. number of clips
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* Do your results indicate the presence of a force other than the gravitational and buoyant forces acting on the balloon?
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No. Given the explanation above in the problem, the only forces that I can determine are acting on the balloon are gravity and the buoyant force.
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`q004. When a certain object coasts up an incline its acceleration has magnitude 100 cm/s^2 and is directed down the incline. When it coasts down the incline its acceleration is 50 cm/s^2 and is directed up the incline. Only gravitational, normal and frictional forces are present.
* Sketch a figure depicting the forces on the object as it coasts up, and as it coasts down the incline. Describe your sketch.
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In the 1st figure of the car going up the incline we have the following:
Normal force perpendicular to incline
Gravity acting down
Motion is going up the incline so frictional force is going opposite the incline
In the 2nd figure all is the same except the frictional force is going up the incline of down the incline because it is the opposite of motion.
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* Are the magnitudes of the force vector depicted in your sketch consistent with the given accelerations? If not, make another sketch and adjust the vectors as necessary. Then describe why you think your sketch is a reasonable representation of the system.
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Going up incline:
Yes, because the vector representing the normal force is 10 times as short as the vector representing the force of gravity.
Going down incline:
Yes, because the vector representing the normal force is 20 times as short as the vector representing the force of gravity.
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* From the given information you can determine the coefficient of friction. You may assume that the normal force varies little in magnitude from the weight of the object. What is the coefficient of friction?
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Going up the incline
coefficient of friction=acceleration of car/acceleration of gravity
=(100cm/s^2)/(980cm/s^2)
=.10
Going down the incline
(50cm/s^2)/(980cm/s^2)
=.05
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* Having determined the coefficient of friction, you can also determine the slope of the incline. For simplicity you can assume that since the slope is small, the magnitude of the weight component parallel to the incline is equal to the slope multiplied by the weight of the object. What do you get?
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This part gave me some trouble. I’m not sure how to find the weight of the object just given the acceleration and the coefficient of friction. Also, how do I find the slope of the incline when also given the coefficient of friction and the acceleration?
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`q005. If parts of the preceding problem gave you trouble, consider an object on an incline with slope .05 and coefficient of friction .03. You can again assume that since the slope is small, the magnitude of the weight component parallel to the incline is equal to the slope multiplied by the weight of the object, and also that the normal force does not differ significantly in magnitude from the object's weight. Let m stand for the mass of the object, g for the acceleration of gravity.
In terms of m and g:
* What is the weight of the object?
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wt=m*g
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* What is the magnitude of its weight parallel to the incline?
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||wt||=||m*.05||
magnitude of weight would be || m g * .05 ||
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* What is the magnitude of the normal force?
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The normal force should be opposite that of gravity shouldn’t it? So, Fnormal=g
g is an acceleration, not a force; the force of gravity would be m g
the normal force is perpendicular to the incline, and in this case is equal and opposite to the component of the weight perpendicular to the incline
in this case, for a small angle, the magnitude of the normal force would be approximately, though not exactly, equal to m g
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* What is the magnitude of the frictional force?
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Ffric=
this would be .03, or 3%, of the normal force
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* What is the magnitude of the net force when the object coasts up the incline?
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* What is the magnitude of the net force when the object coasts down the incline?
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* What therefore are the object's accelerations up, and down, the incline?
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* What are those accelerations in cm/s^2?
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&#Good responses. Let me know if you have questions. &#
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